A 100.0 mL sample of a solution that is 0.125 M in HClO and 0.130 M in NaClO has added to it 2.00 mL of 0.100 M HCl. The Ka for HClO is 2.9 x 10-8. What is the pH of the resulting solution?

To start, I found the moles of each compound, then I added the moles of HClO and HCl to get my total moles of acid. Finally, I plugged it into the Henderson-Hasselbach equation.

I found that 0.0002 moles of HCl was added to a solution that contained 0.0125 moles of HClO and 0.013 moles of NaClO

So...

-log(2.9 -log(2.9*10^-8) + log(0.013/0.0127)
pH=7.55

Does this seem right? I wasn't sure how to approach this problem and I am unsure about the answer I got.

Thanks for any help!

1 Answer
Jul 7, 2018

"pH" = 7.554

Explanation:

Start by finding the quantity of each species in the "NaClO" buffer immediately after the addition of "HCl".

  • n("HCl") = c * V = 2.00 xx 10^(-4) color(white)(l) mol
  • n("HClO") = c * V = 1.25 xx 10^(-2) color(white)(l) mol
  • n("NaClO") = c * V = 1.30 xx 10^(-2) color(white)(l) mol

Hydrochloric acid "HCl" would react with "NaClO" to produce a weak electrolyte "HClO" through the following double decomposition reaction:

stackrel((color(grey)("strong")))("HCl"(aq)) + "NaClO" (aq) color(navy)(to) stackrel((color(grey)("weak")))("HClO"(aq)) + "NaCl"(aq)

"NaClO" is in excess and therefore

n("HClO","final") = n("HClO","initial")+ n("HCl")
color(white)(n("HClO","final")) = 1.2502 xx 10^(-2) color(white)(l) mol

n("NaClO", "final") = n("NaClO", "initial") - n("HCl")
color(white)(n("NaClO","final")) = 1.2998 xx 10^(-2) color(white)(l) mol

The rest of "NaClO" completely disassociates to produce "Na"^(+) and "ClO"^(-).

"NaClO"(s) color(navy)(to) "Na"^(+)(aq) + "ClO"^(-)(aq)

n("ClO"^(-)) = n("NaClO") = 1.30 xx 10^(-2) color(white)(l) mol

The hydrolysis of aqueous "ClO"^(-) from "NaClO" might pose an influence on the acidity (in other words, "pH") of the solution. A "RICE" table can help with the derivation of the equilibrium proton concentration due to this reversible process.

R " ClO"^(-)(aq) + "H"_2"O"(l) color(darkgreen)(rightleftharpoons) "OH"^(-) (aq) + "HClO"(aq)
I color(purple)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------) color(white)(ll)1.2502 xx 10^(-2)
C color(darkgreen)(" "-x color(white)(-------) +x color(white)(---) +x)
E color(navy)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------)1.2502 xx 10^(-2))
color(navy)(color(white)(--) -x color(white)(-----lll-) +x color(white)(---)+x)
(all values in this table are in mol)

K_a = 2.9 xx 10^(-8) for "HClO" such that
K_b = (1.0 xx 10^(-14))/ (2.9 xx 10^(-8)) = 3.448 xx 10^(-7) for "ClO"^(-), its conjugate base.

(n("OH"^(-)) * n("HClO")) / (n("ClO"^(-))) = (["OH"^(-)] * ["HClO"]) / (["ClO"^(-)]) * V= K_b * V = 3.4 xx 10^(-7) * 0.102

Where n("OH"^(-)), n("HClO"), and n("ClO"^(-)) the color(navy)("quantities") (in number of moles) of each of the respective species present in the solution at its color(navy)("equilibrium") position. With reference to the RICE table,

((x) * (1.2502 xx 10^(-2) + x)) / (1.2998 xx 10^(-2) - x) = 3.448 xx 10^(-7) * 0.102

Solving this equation for x yields x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol). Therefore n("OH"^(-)) = x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol) of "OH"^(-) would eventually be present in the solution as it comes to its equilibrium state.

The final solution has a volume of 100.0 + 2.00 = 102.0 color(white)(l) ml (assuming no volume change due to the mixing.) Thus ["OH"^(-)] = n / V = 3.584 xx 10^(-7) color(white)(l) mol at equilibrium.

"pOH" = -log["OH"^(-)] = 6.445
and therefore
"pH" = "pKw" - "pOH" = 7.554.